2003 AIME II Problems/Problem 8: Difference between revisions

Revision as of 19:40, 4 July 2013

Problem

Find the eighth term of the sequence $1440,$ $1716,$ $1848,\ldots,$ whose terms are formed by multiplying the corresponding terms of two arithmetic sequences.

Solution

Solution 1

If you multiply the corresponding terms of two arithmetic sequences, you get the terms of a quadratic function. Thus, we have a quadratic $ax^2+bx+c$ such that $f(1)=1440$ , $f(2)=1716$ , and $f(3)=1848$ . Plugging in the values for x gives us a system of three equations:

$a+b+c=1440$

$4a+2b+c=1716$

$9a+3b+c=1848$

Solving gives $a=-72, b=492,$ and $c=1020$ . Thus, the answer is $-72(8)^2+492\cdot8+1020=348$ .

Solution 2

Setting one of the sequences as $a+nr_1$ and the other as $b+nr_2$ , we can set up the following equalities

$ab = 1440$

$(a+r_1)(b+r_2)=1716$

$(a+2r_1)(b+2r_2)=1848$

We want to find $(a+7r_1)(b+7r_2)$

Foiling out the two above, we have

$ab + ar_2 + br_1 + r_1r_2 = 1716$ and $ab + 2ar_2 + 2br_1 + 4r_1r_2 = 1848$

Plugging in $ab=1440$ and bringing the constant over yields

$ar_2 + br_1 + r_1r_2 = 276$

$ar_2 + br_1 + 2r_1r_2 = 204$

Subtracting the two yields $r_1r_2 = -72$ and plugging that back in yields $ar_2 + br_1 = 348$

Now we find

$(a+7r_1)(b+7r_2) = ab + 7(ar_2 + br_1) + 49r_1r_2 = 1440 + 7(348) + 49(-72) = \boxed{348}$

@@ Line 43: / Line 43: @@
 == See also ==
 {{AIME box|year=2003|n=II|num-b=7|num-a=9}}
+{{MAA Notice}}

2003 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 7	Followed by Problem 9
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

Page

Toolbox

Search